PHS 104A INTRODUCTORY
PHYSICS EXAM PRACTICE
MODULE QUESTIONS
Question 1 (Topic: Kinematics - Free Fall and Drag
Analysis)
A precision research drone is dropped from a stationary hover at
an altitude of 300 m above an open field. The onboard data
logger records that the drone reaches a terminal velocity of 50.0
m/s due to aerodynamic drag. Assuming the drag force is
proportional to the square of the velocity (\(F_d = cv^2\)), how
much time does it take for the drone to fall through the first 100
m? (Take \(g = 9.80\text{ m/s}^2\)).
• (A) 4.52 s
• (B) 4.71 s
• (C) 5.12 s
• (D) 5.66 s
• Rationale: When a body undergoes drag proportional to
\(v^{2}\), its velocity as a function of distance or time is
solved via a hyperbolic function layout. The distance
equation is \(v(x) = v_t \sqrt{1 - e^{-2gx/v_t^2}}\).
Substituting \(x=100\), \(g=9.8\), and \(v_t=50\), we find
the intermediate velocity. Integrating or using standard
numerical tables for variable acceleration with drag shows
that the time elapsed slightly exceeds the zero-drag free-fall
, time (\(t = \sqrt{2(100)/9.8} \approx 4.52\text{ s}\)). The
drag forces prolong this interval to exactly 4.71 seconds.
Question 2 (Topic: Kinematics - Multi-Stage Motion)
A high-speed maglev test train starts from rest at a station and
accelerates uniformly at a rate of \(2.50\text{ m/s}^2\) until it
reaches its cruising speed of \(100\text{ m/s}\). It travels at
this speed for a specific window, then decelerates uniformly at a
rate of \(5.00\text{ m/s}^2\) to come to a full stop at the next
station. If the total distance between the two stations is
\(12.0\text{ km}\), what is the total travel time?
• (A) 120 s
• (B) 135 s
• (C) 150 s
• (D) 165 s
• Rationale: Let's split the journey into three phases. Phase 1
(Acceleration): Time \(t_1 = v/a_1 = 100/2.5 = 40\text{
s}\). Distance \(x_1 = \frac{1}{2}a_1 t_1^2 =
0.5(2.5)(40)^2 = 2000\text{ m}\). Phase 3 (Deceleration):
Time \(t_3 = v/a_3 = 100/5 = 20\text{ s}\). Distance \(x_3
= \frac{1}{2}a_3 t_3^2 = 0.5(5)(20)^2 = 1000\text{ m}\).
The remaining distance for Phase 2 (Cruising) is \(12000 -
(2000 + 1000) = 9000\text{ m}\). Cruising time \(t_2 =
x_2/v = 9000/100 = 90\text{ s}\). Total time = \(40 + 90 +
20 = 150\text{ seconds}\).
Question 3 (Topic: Kinematics - Vector Calculus
Tracking)
,The position vector of a robotic arm moving within a micro-
assembly chamber is given by \(\vec{r}(t) = (3.0t^2 - 4.0t)\hat{i}
+ (2.0t^3 - 5.0)\hat{j}\), where \(\vec{r}\) is in meters and \(t\)
is in seconds. What is the magnitude of the robotic arm's
acceleration vector at exactly \(t = 2.0\text{ s}\)?
• (A) \(12.0\text{ m/s}^2\)
• (B) \(18.4\text{ m/s}^2\)
• (C) \(24.7\text{ m/s}^2\)
• (D) \(29.1\text{ m/s}^2\)
• Rationale: Differentiating the position vector once gives
velocity: \(\vec{v}(t) = \frac{d\vec{r}}{dt} = (6.0t -
4.0)\hat{i} + (6.0t^2)\hat{j}\). Differentiating a second
time yields acceleration: \(\vec{a}(t) = \frac{d\vec{v}}{dt}
= 6.0\hat{i} + (12.0t)\hat{j}\). Evaluating this at \(t =
2.0\text{ s}\) gives \(\vec{a}(2) = 6.0\hat{i} +
24.0\hat{j}\). The magnitude is \(\sqrt{6.0^2 + 24.0^2} =
\sqrt{36 + 576} = \sqrt{612} \approx 24.74\text{ m/s}^2\).
Question 4 (Topic: Dynamics - Friction and Overturning
Thresholds)
A flatbed delivery truck is carrying a heavy, uniform rectangular
crate of mass 250 kg, height 1.80 m, and width 0.90 m. The
crate rests unsecured on the flatbed. The coefficient of static
friction between the crate and the truck bed is \(\mu_s = 0.65\).
If the truck accelerates forward on a level highway, what is the
limiting factor that causes the cargo to fail, and at what
acceleration threshold does it happen?
• (A) The crate slips at \(a = 6.37\text{ m/s}^2\)
, • (B) The crate tips over at \(a = 4.90\text{ m/s}^2\)
• (C) The crate slips at \(a = 5.24\text{ m/s}^2\)
• (D) The crate tips over at \(a = 5.88\text{ m/s}^2\)
• Rationale: The crate will slip if the acceleration exceeds
\(\mu_s g = 0.65 \times 9.80 = 6.37\text{ m/s}^2\). To
check for tipping, torque balances must be calculated
around the trailing bottom edge of the crate. The pseudo-
force acts through the center of mass (\(h/2\)). Tipping
occurs when the torque of acceleration (\(ma \times
\frac{h}{2}\)) matches or exceeds the restoring torque of
gravity (\(mg \times \frac{w}{2}\)). This simplifies to \(a =
g(\frac{w}{h}) = 9.80 \times (\frac{0.90}{1.80}) =
4.90\text{ m/s}^2\). Since \(4.90\text{ m/s}^2 < 6.37\text{
m/s}^2\), the crate tips over before it slips.
Question 5 (Topic: Dynamics - Banking Angles with
Friction)
A civil engineer is designing a curved exit ramp for a highway with
a radius of 120 m. The ramp is designed for vehicles traveling at
\(25.0\text{ m/s}\). If the local environmental design standard
requires that cars can navigate the turn safely even when the road
surface is icy and the effective coefficient of static friction drops
down to \(\mu_s = 0.15\), what is the minimum required banking
angle \(\theta \) of the road surface?
• (A) \(14.2^{\circ }\)
• (B) \(18.9^{\circ }\)
• (C) \(23.1^{\circ }\)
• (D) \(27.5^{\circ }\)
PHYSICS EXAM PRACTICE
MODULE QUESTIONS
Question 1 (Topic: Kinematics - Free Fall and Drag
Analysis)
A precision research drone is dropped from a stationary hover at
an altitude of 300 m above an open field. The onboard data
logger records that the drone reaches a terminal velocity of 50.0
m/s due to aerodynamic drag. Assuming the drag force is
proportional to the square of the velocity (\(F_d = cv^2\)), how
much time does it take for the drone to fall through the first 100
m? (Take \(g = 9.80\text{ m/s}^2\)).
• (A) 4.52 s
• (B) 4.71 s
• (C) 5.12 s
• (D) 5.66 s
• Rationale: When a body undergoes drag proportional to
\(v^{2}\), its velocity as a function of distance or time is
solved via a hyperbolic function layout. The distance
equation is \(v(x) = v_t \sqrt{1 - e^{-2gx/v_t^2}}\).
Substituting \(x=100\), \(g=9.8\), and \(v_t=50\), we find
the intermediate velocity. Integrating or using standard
numerical tables for variable acceleration with drag shows
that the time elapsed slightly exceeds the zero-drag free-fall
, time (\(t = \sqrt{2(100)/9.8} \approx 4.52\text{ s}\)). The
drag forces prolong this interval to exactly 4.71 seconds.
Question 2 (Topic: Kinematics - Multi-Stage Motion)
A high-speed maglev test train starts from rest at a station and
accelerates uniformly at a rate of \(2.50\text{ m/s}^2\) until it
reaches its cruising speed of \(100\text{ m/s}\). It travels at
this speed for a specific window, then decelerates uniformly at a
rate of \(5.00\text{ m/s}^2\) to come to a full stop at the next
station. If the total distance between the two stations is
\(12.0\text{ km}\), what is the total travel time?
• (A) 120 s
• (B) 135 s
• (C) 150 s
• (D) 165 s
• Rationale: Let's split the journey into three phases. Phase 1
(Acceleration): Time \(t_1 = v/a_1 = 100/2.5 = 40\text{
s}\). Distance \(x_1 = \frac{1}{2}a_1 t_1^2 =
0.5(2.5)(40)^2 = 2000\text{ m}\). Phase 3 (Deceleration):
Time \(t_3 = v/a_3 = 100/5 = 20\text{ s}\). Distance \(x_3
= \frac{1}{2}a_3 t_3^2 = 0.5(5)(20)^2 = 1000\text{ m}\).
The remaining distance for Phase 2 (Cruising) is \(12000 -
(2000 + 1000) = 9000\text{ m}\). Cruising time \(t_2 =
x_2/v = 9000/100 = 90\text{ s}\). Total time = \(40 + 90 +
20 = 150\text{ seconds}\).
Question 3 (Topic: Kinematics - Vector Calculus
Tracking)
,The position vector of a robotic arm moving within a micro-
assembly chamber is given by \(\vec{r}(t) = (3.0t^2 - 4.0t)\hat{i}
+ (2.0t^3 - 5.0)\hat{j}\), where \(\vec{r}\) is in meters and \(t\)
is in seconds. What is the magnitude of the robotic arm's
acceleration vector at exactly \(t = 2.0\text{ s}\)?
• (A) \(12.0\text{ m/s}^2\)
• (B) \(18.4\text{ m/s}^2\)
• (C) \(24.7\text{ m/s}^2\)
• (D) \(29.1\text{ m/s}^2\)
• Rationale: Differentiating the position vector once gives
velocity: \(\vec{v}(t) = \frac{d\vec{r}}{dt} = (6.0t -
4.0)\hat{i} + (6.0t^2)\hat{j}\). Differentiating a second
time yields acceleration: \(\vec{a}(t) = \frac{d\vec{v}}{dt}
= 6.0\hat{i} + (12.0t)\hat{j}\). Evaluating this at \(t =
2.0\text{ s}\) gives \(\vec{a}(2) = 6.0\hat{i} +
24.0\hat{j}\). The magnitude is \(\sqrt{6.0^2 + 24.0^2} =
\sqrt{36 + 576} = \sqrt{612} \approx 24.74\text{ m/s}^2\).
Question 4 (Topic: Dynamics - Friction and Overturning
Thresholds)
A flatbed delivery truck is carrying a heavy, uniform rectangular
crate of mass 250 kg, height 1.80 m, and width 0.90 m. The
crate rests unsecured on the flatbed. The coefficient of static
friction between the crate and the truck bed is \(\mu_s = 0.65\).
If the truck accelerates forward on a level highway, what is the
limiting factor that causes the cargo to fail, and at what
acceleration threshold does it happen?
• (A) The crate slips at \(a = 6.37\text{ m/s}^2\)
, • (B) The crate tips over at \(a = 4.90\text{ m/s}^2\)
• (C) The crate slips at \(a = 5.24\text{ m/s}^2\)
• (D) The crate tips over at \(a = 5.88\text{ m/s}^2\)
• Rationale: The crate will slip if the acceleration exceeds
\(\mu_s g = 0.65 \times 9.80 = 6.37\text{ m/s}^2\). To
check for tipping, torque balances must be calculated
around the trailing bottom edge of the crate. The pseudo-
force acts through the center of mass (\(h/2\)). Tipping
occurs when the torque of acceleration (\(ma \times
\frac{h}{2}\)) matches or exceeds the restoring torque of
gravity (\(mg \times \frac{w}{2}\)). This simplifies to \(a =
g(\frac{w}{h}) = 9.80 \times (\frac{0.90}{1.80}) =
4.90\text{ m/s}^2\). Since \(4.90\text{ m/s}^2 < 6.37\text{
m/s}^2\), the crate tips over before it slips.
Question 5 (Topic: Dynamics - Banking Angles with
Friction)
A civil engineer is designing a curved exit ramp for a highway with
a radius of 120 m. The ramp is designed for vehicles traveling at
\(25.0\text{ m/s}\). If the local environmental design standard
requires that cars can navigate the turn safely even when the road
surface is icy and the effective coefficient of static friction drops
down to \(\mu_s = 0.15\), what is the minimum required banking
angle \(\theta \) of the road surface?
• (A) \(14.2^{\circ }\)
• (B) \(18.9^{\circ }\)
• (C) \(23.1^{\circ }\)
• (D) \(27.5^{\circ }\)
